Question: $f(r)=-(r-12)(r+3)$ 1) What are the zeros of the function? Write the smaller $r$ first, and the larger $r$ second. $\text{smaller }r=$
$\begin{aligned} -(r-12)(r+3)&=0 \\\\ r-12=0&\text{ or }r+3=0 \\\\ r={12}&\text{ or }r={-3} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $r$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }r\text{-coordinate}&=\dfrac{({12})+({-3})}{2} \\\\ &={\dfrac{9}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $f\left({\dfrac{9}{2}}\right)$ : $\begin{aligned} f\left({\dfrac{9}{2}}\right)&=-\left({\dfrac{9}{2}}-12\right)\left({\dfrac{9}{2}}+3\right) \\\\ &=-\left(-\dfrac{15}{2}\right)\left(\dfrac{15}{2}\right) \\\\ &=\dfrac{225}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }r&=-3 \\\\ \text{larger }r&=12 \end{aligned}$ The vertex of the parabola is at $\left(\dfrac{9}{2},\dfrac{225}{4}\right)$